Pls ! : ) based on valence electron configurations, which of the following is the accurate formula for a compound formed from sodium (na) and oxygen (: all numerical values below are subscripts) *

The study of chemicals is called chemistry. There are two types of solution in the chemistry and these are acid and base.

The correct answer is

What are moles?

Thus sodium needs to lose one electron in order to attain noble gas configuration and Oxygen needs to gain two electrons to attain noble gas configuration.

Thus Na will lose one electron to form and Oxygen gain two electrons to form In order to form a stable neutral compound, two sodium ions are required.

Thus the formula will be .

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The accurate formula is .

Explanation: Atomic no of Sodium is 11. The electronic configuration is:

Na:

Atomic no of Oxygen is 8. The electronic configuration is:

O:

Thus sodium needs to lose one electron in order to attain noble gas configuration and Oxygen needs to gain two electrons to attain noble gas configuration.

Thus Na will lose one electron to form and Oxygen gain two electrons to form . In order to form a stable neutral compound, two sodium ions are required.

Thus the formula will be .

See explanation below

Explanation:

You are missing certain data. I found one, in the attached picture. All you have to do is replace your data in this procedure to get an accurate answer.

According to the picture attached, we have 2.4 g of HBr and 1.9 g of NaOH. This mix produces 0.411 g of water. To get the %yield of water, we first need to see how many theorical grams of water are formed. With this we can calculate the %yield.

The overall reaction is:

HBr + NaOH --------> NaBr + H₂O

We have a mole ratio of 1:1 so, let's see which is the limiting reactant between the acid and the base, using the reported molar mass for each (MM HBr = 80.9119 g/mol; MM NaOH = 39.997 g/mol)

moles HBr = 2.4 / 80.9119 = 0.02966 moles

moles NaOH = 1.9 / 39.997 = 0.0475 moles

So, we can clearly see that the moles of NaOH are in excess, so the HBr is the limiting reactant.

As we have a mole ratio of 1:1 with each compound here, and that the HBr is the limiting reactant we can say that:

moles HBr = moles H₂O = 0.02966 moles of water

Now, using the molar mass of water, let's calculate the theorical yield of water: (MM water = 18.01528 g/mol

m H₂O = 0.02966 * 18.01528 = 0.534 g

We have the theorical yield, let's calculate the % yield:

% = 0.411 / 0.534 * 100

% = 76.92%